SATHEE: Differential Equations 1 Question 15

Differential Equations 1 Question 15

15. Let $f : R \to R$ be a differentiable function with $f (0) = 0$ . If $y = f (x)$ satisfies the differential equation $\frac{d y}{d x} = (2 + 5 y) (5 y - 2)$ , then the value of $lim_{x \to - \infty} f (x)$ is $\dots \dots . . . .)$

Assertion and Reason

For the following question, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.

(a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I.

(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I.

(d) Statement I is false; Statement II is true.

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Answer:

Correct Answer: 15. (0)

Solution:

We have,

$\begin{aligned} \frac{d y}{d x} = (2 + 5 y) (5 y - 2) \\ \Rightarrow \frac{d y}{25 y^{2} - 4} = d x \Rightarrow \frac{1}{25} \frac{d y}{y^{2} - \frac{4}{25}} = d x \end{aligned}$

On integrating both sides, we get

$\begin{aligned} \frac{1}{25} \int \frac{d y}{y^{2} - \frac{2}{5}} = \int d x \\ \Rightarrow & \frac{1}{25} \times \frac{1}{2 \times \frac{2}{5}} \log | \frac{y - 2 / 5}{y + 2 / 5} | = x + C \\ \Rightarrow & \log | \frac{5 y - 2}{5 y + 2} | = 20 (x + C) \\ \Rightarrow & | \frac{5 y - 2}{5 y + 2} | = A e^{20 x} [∵ e^{20 C} = A] \end{aligned}$

when $x = 0 \Rightarrow y = 0$ , then $A = 1$

$∴ | \frac{5 y - 2}{5 y + 2} | = e^{20 x}$

$lim_{x \to - \infty} | \frac{5 f (x) - 2}{5 f (x) + 2} | = lim_{x \to - \infty} e^{20 x}$

$\begin{aligned} \Rightarrow lim_{n \to - \infty} | \frac{5 f (x) - 2}{5 f (x) + 2} | = 0 \\ \Rightarrow lim_{n \to - \infty} 5 f (x) - 2 = 0 \\ \Rightarrow lim_{n \to - \infty} f (x) = \frac{2}{5} = 0.4 \end{aligned}$